MATHEMATICS COLLEGE

The state department of transportation is coordinating road crews to fix potholes after a particularly snowy winter. Initial estimates gave an average of 7.8 potholes per mile of highway. Find the probability that there are 11 or more potholes in a randomly selected one-mile stretch of highway. Use Excel to find the probability.

Answers

Answer 1

Answer:

P(X≥11) = 0.1648

Step-by-step explanation:

Given

Mean, μ = 7.3

Potholes, n = 11

The interpretation of the question is to calculate P(X ≥ 11)

It is known that

P(X) = P(1) + P(2) +.......+P(infinite)

We can say that

P(X) = P(X≤10) + P(X>10)

Make P(X>10) the subject of formula

P(X>10) = P(X) - P(X≤10)

P(X>10) is equivalent to P(X≥11) and P(X) = 1.

By substituton, we have

P(X≥11) = 1 - P(X≤10)

So, we'll solve P(X≤10) using the following steps using n as 10.

To solve the above question using Microsoft Office Excel, follow the highlighted steps below

1. First goto FORMULAS tan

2. Select INSERT FUNCTION.

3. Select the POISSON.DIST function.

4. Enter the values for the number of events and the mean of occurrences per interval. In this case, enter 10 and 7.8, in that order and 1 for Cumulative since this is a cumulative probability.

5. Press OK.

Excel would display the probability.

In this case, it is 0.83523

Remember that

P(X≥11) = 1 - P(X≤10)

By substituton

P(X≥11) = 1 - 0.83523

P(X≥11) = 0.16477

Approximately,

P(X≥11) = 0.1648

(See attachment)

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The marks on a combination lock are numbered 0 to 39. If the lock is at mark 19, and the dial is turned one mark clockwise, it will be at mark 18. If the lock is at mark 19 and turned 137 marks clockwise, at what mark will it be

Answers

Answer:

Mark 2

Step-by-step explanation:

There are 40 marks on the lock.

Every full turn of 40 marks, the lock will come back to mark 19.

137/40 = 3R17

You can make three full turns (which is equivalent to doing nothing), and then you must move 17 more marks.

19 - 17 = 2

The lock will be at mark 2.

If the lock is at mark 19 and turned 137 marks clockwise, it will be at mark 2.

What is combination lock?

Combination lock is opened by rotating a set of number dials or marks graduated in a sequence.

The marks on a combination lock are numbered 0 to 39. If the lock is at mark 19, and the dial is turned one mark clockwise, it will be at mark 18. If the lock is at mark 19 and turned 137 marks clockwise,

137 = 40a +b where b<40

Divide 137 by 40, so we get the quotient 3 i.e., a =3.

Substitute a = 3 , the value of b is

b = 17

If the lock is at mark 19 and turned 137 marks clockwise, it will be at mark,

19-17 = 2

Thus, If the lock is at mark 19 and turned 137 marks clockwise, it will be at mark 2.

Learn more about combination lock.

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Which ordered pair is a solution of the equation? y=-2x+5y=−2x+5 (Choice A) A Only (2,-9)(2,−9) (Choice B) B Only (-2,9)(−2,9) (Choice C) C Both (2,-9)(2,−9) and (-2,9)(−2,9) (Choice D) D Neither

Answers

Answer:

Choice B: Only (-2, 9)

Step-by-step explanation:

Of the two choices, only the point (-2, 9) satisfies the equation:

... y = -2x +5

... 9 = -2(-2) +5 = 4 +5 = 9

Answer:

The correct answer is B. Only (-2,9) is a solution of the equation.

Step-by-step explanation:

To figure out if an ordered pair is a solution to an equation, you could perform a test.

Identify the x-value in the ordered pair and plug it into the equation. When you simplify, if the y-value you get is the same as the y-value in the ordered pair, then that ordered pair is indeed a solution to the equation.

We have the equation

$y=-2x+5$

and two different ordered pairs (2, -9) and (-2, 9).

We use the x-value in the first ordered pair, x = 2, and we plug it into the equation.

$y=-2(2)+5\ny=-4+5\ny=1$

If we compare with the ordered pair (2, -9), y must be equal to -9, but when we plugged x = 2 into the equation, we obtained that y = 1, so this is not a solution.

Using the x-value of the second ordered pair, x = -2, we get

$y=-2(-2)+5\ny=4+5\ny=9$

Because we obtained y = 9, when we plugged x = -2, the ordered pair (-2, 9) is a solution of the equation.

The correct answer is B. Only (-2,9) is a solution of the equation.

The temperature, H, in °F, of a cup of coffee t hours after it is set out to cool is given by the following equation. H = 70 + 120(1/4)t (a) What is the coffee's temperature initially (that is, at time t = 0)? 190 °F What is the coffee's temperature after 1 hour? 100 °F What is the coffee's temperature after 2 hours? (Round your answer to one decimal place.) 2 °F (b) How long does it take the coffee to cool down to 85°F? (Round your answer to three decimal places.) 5 hr How long does it take the coffee to cool down to 75°F? (Round your answer to three decimal places.) 5 hr

Answers

Answer:

The temperature a t = 0 is 190 °F

The temperature a t = 1 is 100 °F

The temperature a t = 2 is 77.5 °F

It takes 1.5 hours to take the coffee to cool down to 85°F

It takes 2.293 hours to take the coffee to cool down to 75°F

Step-by-step explanation:

We know that the temperature in °F, of a cup of coffee t hours after it is set out to cool is given by the following equation:

$H(t)=70+120((1)/(4))^t$

a) To find the temperature a t = 0 you need to replace the time in the equation:

$H(0)=70+120((1)/(4))^0\nH(0)=70+120\cdot 1\nH(0) = 70+120\nH(0)=190 \:\°F$

b) To find the temperature after 1 hour you need to:

$H(1)=70+120((1)/(4))^1\nH(1)=70+120((1)/(4))\nH(1) = 70+30\nH(1)=100 \:\°F$

c) To find the temperature after 2 hours you need to:

$H(2)=70+120((1)/(4))^2\nH(2)=70+120((1)/(16))\nH(2) = 70+(15)/(2) \nH(2)=77.5 \:\°F$

d) To find the time to take the coffee to cool down $85 \:\°F$ , you need to:

$85 = 70+120((1)/(4))^t\n70+120\left((1)/(4)\right)^t=85\n70+120\left((1)/(4)\right)^t-70=85-70\n120\left((1)/(4)\right)^t=15\n(120\left((1)/(4)\right)^t)/(120)=(15)/(120)\n\left((1)/(4)\right)^t=(1)/(8)$

$\mathrm{If\:}f\left(x\right)=g\left(x\right)\mathrm{,\:then\:}\ln \left(f\left(x\right)\right)=\ln \left(g\left(x\right)\right)$

$\ln \left(\left((1)/(4)\right)^t\right)=\ln \left((1)/(8)\right)$

$\mathrm{Apply\:log\:rule}=\log _a\left(x^b\right)=b\cdot \log _a\left(x\right)$

$t\ln \left((1)/(4)\right)=\ln \left((1)/(8)\right)$

$t=(\ln \left((1)/(8)\right))/(\ln \left((1)/(4)\right))\nt=(3)/(2) = 1.5 \:hours$

e) To find the time to take the coffee to cool down $75 \:\°F$ , you need to:

$75=70+120\left((1)/(4)\right)^t\n70+120\left((1)/(4)\right)^t=75\n70+120\left((1)/(4)\right)^t-70=75-70\n120\left((1)/(4)\right)^t=5\n\left((1)/(4)\right)^t=(1)/(24)$

$\ln \left(\left((1)/(4)\right)^t\right)=\ln \left((1)/(24)\right)\nt\ln \left((1)/(4)\right)=\ln \left((1)/(24)\right)\nt=(\ln \left(24\right))/(2\ln \left(2\right)) \approx = 2.293 \:hours$

What is the answer to (2a^4/b^5)^3

Answers

I hope this helps you

Mr. Matinews and Mr. Peters are scuba diving. Mr. Matthews started out 12 feet belowthe surface. He descended 8 feet, rose 7 feet, and descended 13 more feet. Then he
rested. Mr. Peters started out at the surface. He descended 25 feet, rose 8 feet and
descended another 6 feet. Then he rested. Which person rested at a greater depth?
Annotations
Organized workspace
Check work/answer

Answers

Answer:

Mr.Matthews

Step-by-step explanation:

Mr.Matthews: Started at -12

-12 - 8 + 7 - 13 = -20 + 7 - 13 = -13 - 13 = -26

Mr. Peters: Started at -25

-25 + 8 - 6 = -17 - 6 = -23

In this case, -26 is greater than -23.

Mr.Matthews rested at a greater depth.

Hope this helps :)

In how many ways can a subcommittee of 6 students be chosen from a committee which consists of 10 senior members and 12 junior members if the team must consist of 4 senior members and 2 junior members?